Equivariant Map of Group Action
Definition1 2
Suppose a group $G$, and two actions $\ast_{1} : G \times X \to X$, $\ast_{2} : G \times Y \to Y$ are given. A function $f : X \to Y$ between two $G$-sets is said to be equivariant concerning $G$ if it satisfies the following condition.
$$ f(g \ast_{1} x) = g \ast_{2} f(x), \qquad \forall g \in G, x \in X $$
Explanation
Simply speaking, the result of applying the transformation in the domain before applying the function is the same as applying the function first and then applying the transformation in the codomain.
Example
Consider the following example:
- $G = \braket{\mathbb{Z}, +}$ the group of integer addition
- $X = \mathbb{Z}$ the set of integers
- $Y = \left\{ 1, 0 \right\}$
- $\ast_{1} : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ is addition $g \ast_{1} x = g + x$
- $\ast_{2} : \mathbb{Z} \times \left\{ 1, 0 \right\} \to \left\{ 1, 0 \right\}$ is the following action: if $g$ is even, then $y$ is unchanged; if $g$ is odd, then $y$ is inverted. $$ g \ast_{2} y = \begin{cases} y & \text{if } g \text{ is even} \\ 1 - y & \text{if } g \text{ is odd} \end{cases} $$
- $f : \mathbb{Z} \to \left\{ 1, 0 \right\}$ is a function that sends even numbers to $1$, and odd numbers to $0$.
Then, the following holds.
$$ f(g \ast_{1} x) = g \ast_{2} f(x), \qquad \forall g \in \mathbb{Z}, x \in \mathbb{Z} $$